An average force of 125,N is applied on a mac | vedclass

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(B) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets.$F = \frac{dp}{dt} = n \cdot m \cdot v$where $n$ is

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$50$

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(B) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets.$F = \frac{dp}{dt} = n \cdot m \cdot v$where $n$ is the number of bullets fired per second,$m$ is the mass of each bullet,and $v$ is the velocity of the bullets.Given:Force $F = 125\,N$Mass $m = 10\,g = 10 	imes 10^{-3}\,kg = 0.01\,kg$Velocity $v = 250\,m/s$Substituting the values into the formula:$125 = n 	imes 0.01 	imes 250$$125 = n 	imes 2.5$$n = \frac{125}{2.5} = 50$Therefore,the number of bullets fired per second is $50$.

An average force of $125\,N$ is applied on a machine gun firing bullets each of mass $10\,g$ at the speed of $250\,m/s$ to keep it in position. The number of bullets fired per second by the machine gun is:

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